Question

A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground.

Answer 1

$h=40m,u=50m/sec$

Let the speed be 'v' when it strikes the ground.

Applying law of conservation of energy,

$mgh+\frac{1}{2}m{u}^2\frac{1}{2}m{v}^2$

$\Rightarrow 10\times 40+\left(\frac{1}{2}\right)\times 2500=\frac{1}{2}{v}^2$

$\Rightarrow v^2=3300$

$\Rightarrow v=57.4m/sec=58m/sec$